2001 AMC 10 Problems/Problem 2

Problem

A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$\textbf{(A) }\ -4\le x\le -2\qquad\textbf{(B) }\ -2 < x\le 0\qquad\textbf{(C) }0$ $< x \le 2 \qquad \textbf{(D) }\ 2 < x\le 4\qquad\textbf{(E) }\ 4 < x\le 6$

Solution

We can write our equation as $x= \left(\frac{1}{x} \right) \cdot (-x) +2 = -1+2 = 1$ . Therefore, $\boxed{\textbf{(C) }0 < x\le 2}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2001 AMC 10 Problems/Problem 2

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